'bertoldo' created a special Christmas puzzle, a "prime numbers only" puzzle (!)

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Thanks, bertoldo, for prepearing this Christmas puzzle . It's very amused and interesting.

When I sent a similar model (Dec 2011, thread "An easy 6x6 with skipped numbers", in the Section "Specific puzzles / your own puzzles") I was thinking more in the general possibility of using "any" set of "skipped" numbers, but the idea of being only prime numbers, like in this case, though being a particular case of the more general case, is specially curious and interesting (I love the prime numbers). In all cases the first job is always to determine which individual numbers compose the puzzle, right?.

However, this particular example requires some slight modification since it has two different solutions (by swapping numbers 2 and 5 in columns "b" and "c" and consequently swapping all numbers except the 1 in rows 3 and 4 only). But the fact that the solution is not unique does not decrease the beauty of the puzzle anyway.

Edit: Patrick, in my opinion it's not necessary to anticipate that the puzzle is made by only prime numbers since that must be deduced from the puzzle itself (this should be the general case for the "skipped" numbers puzzles)

'bertoldo' created a special Christmas puzzle, a "prime numbers only" puzzle (!)

...

Thanks, bertoldo, for prepearing this Christmas puzzle . It's very amused and interesting.

Agreed! Thanks, bertoldo!

clm wrote:

Edit: Patrick, in my opinion it's not necessary to anticipate that the puzzle is made by only prime numbers since that must be deduced from the puzzle itself (this should be the general case for the "skipped" numbers puzzles)

Some specifics about the puzzle follow. I'll try to stay general but if you want to solve it without my thoughts potentially cluttering the process, skip this next bit.

**skip**In fact, my first pass at the puzzle was using {2,3,5,7,11,13}. I started with row 6 then got bogged down in row 5 before noticing the 3: cage at d1d2 and then the 8+ at 3b3c4b. From there, given that the description specified prime numbers it was clear I needed {1,2,3,5,7,11} and not {1,3,4,5,7,11}. Even without that hint the 9- at a1b1 is the last piece needed to get the correct set.**/skip**

... Some specifics about the puzzle follow. I'll try to stay general but if you want to solve it without my thoughts potentially cluttering the process, skip this next bit.

**skip**In fact, my first pass at the puzzle was using {2,3,5,7,11,13}. I started with row 6 then got bogged down in row 5 before noticing the 3: cage at d1d2 and then the 8+ at 3b3c4b. From there, given that the description specified prime numbers it was clear I needed {1,2,3,5,7,11} and not {1,3,4,5,7,11}. Even without that hint the 9- at a1b1 is the last piece needed to get the correct set.**/skip**

This was a really fun puzzle. Happy holidays to all.

Thanks.Yes, correct, if you do not know in advance that the puzzle is made with prime numbers only, you must analyze all possibilities and that makes the puzzle more amused. A full analysis (to eliminate the possibilities for other big numbers, prime or not) could start considering the pair [1,21] for the cage "21x". This is easy: 5 and 11 are given numbers (individual cells); if the number 21 is present in the puzzle then in row 6 it must be c6 = 21 >>> c5 = 17, and since the 17 is part of the puzzle >>> a6 = 17 (now the only available cell for the 17); now we have 1, 5, 11, 17 and 21 so "10+" = [1,4,5] but with the set [1,4,5,11,17,21] the cage "9-" cann't be satisfied. Consequently "21x" = [3,7].

Next: We now have 3, 5, 7 and 11 defined. Row 6: With a 7 in column "a" either c6 = 7 or the 7 is inside "10+"; in this last case the 1 and 2 come to the final pool [1,2,3,5,7,11]; if c6 = 7 then the 3 of row 6 goes to "10+" along with 2 and 5 being [1,3,6] invalid since [1,3,5,6,7,11] does not satisfy "9-". We have arrived to 2, 3, 5, 7 and 11. We need an additional number and looking to cage "8+" we eliminate [1,1,6] (two new numbers, the 1 and the 6), [2,2,4] because [2,3,4,5,7,11] does not satisfy "3:", [2,3,3] due to a 3 already in "21x", [1,3,4] (two new numbers, the 1 and the 4) so "8+" = [1,2,5] and the 1 is the pending number. The full set is then [1,2,3,5,7,11] and any other set up to the infinite is impossible.

So "3:" = [1,3], etc., until you arrive to the two solutions commented (the difference between these two solutions are some permutations, as commented previously).

Perhaps a more easy explanation in: http://en.wikipedia.org/wiki/Prime_number .Interesting aspect what sheldolina underlines. However, this "philosophical" debate disappears announcing: 'bertoldo' created a special Christmas puzzle, "a prime numbers (plus the numbers 1 and 2) only" puzzle (!) . And In this way we can also extend the discussion to number 2 .

It is true that in a not very far past the 1 was considered a prime (and taught so); in this way, for instance, the Goldbach's conjecture (that every even integer greater than two is the sum of two different primes) looks more consistent: 4 = 3 + 1, 6 = 5 + 1, ... Particularly myself I would prefer to have the 1 among the primes (though science is not a question of preferences I am affraid ) so we could have the "null" number for the multiplication operation included in the "body" of primes and then when we define a prime number as "such a number that can only be divided by itself or the 1", the number 1 itself does not sound as a "strange" in this equation.

But it is also true that nowadays and to be consistent with other theorems it's important to exclude the 1. Apart of this, the properties of prime numbers are fascinanting, as the "twin prime" conjecture (that there are infinitely pairs of prime numbers whose difference is 2), I was taught that commonly in the form of multiple of 6 plus or minus 1, like 5,7; 11,13; 17,19; 29,31; 41,43; 59,61; 71,73, etc..

nicow

Posted on:Mon Dec 30, 2013 1:18 pm

Posts: 66Joined: Fri May 13, 2011 9:37 am

Re: a Christmas puzzle by bertoldo

I agree with 1 being a prime. Why should there be just one 1?

When I got work for the first time, I used an electro-mechanical calculating machine. It made much noise. Negative numbers were printed red. When you repeatedly subtracted 1 from 5, it printed 4,3,2,1,0 in black. When you added 1 to -5, it printed -4,-3,-2,-1,-0 in red. So there were a black 0 and a red 0 !

So 1 can be seen as two different numbers equalling each other. Ha ha.